Question: The ellipse whose equation is
\[\frac{x^2}{25} + \frac{y^2}{9} = 1\]is graphed below.  The chord $\overline{AB}$ passes through a focus $F$ of the ellipse.  If $AF = \frac{3}{2},$ then find $BF.$

[asy]
unitsize (0.6 cm);

pair A, B, F;

F = (4,0);
A = (35/8,3*sqrt(15)/8);
B = (55/16,-9*sqrt(15)/16);

draw(xscale(5)*yscale(3)*Circle((0,0),1));
draw(A--B);
draw((-6,0)--(6,0));
draw((0,-4)--(0,4));

dot("$A$", A, NE);
dot("$B$", B, SE);
dot("$F$", F, NW);
[/asy]
In the given ellipse, $a = 5$ and $b = 3,$ so $c = \sqrt{a^2 - b^2} = 4.$  We can take $F = (4,0).$

Let $A = (x,y).$  Then $\frac{x^2}{25} + \frac{y^2}{9} = 1$ and
\[(x - 4)^2 + y^2 = \left( \frac{3}{2} \right)^2 = \frac{9}{4}.\]Solving for $y^2$ in $\frac{x^2}{25} + \frac{y^2}{9} = 1,$ we get
\[y^2 = \frac{225 - 9x^2}{25}.\]Substituting, we get
\[(x - 4)^2 + \frac{225 - 9x^2}{25} = \frac{9}{4}.\]This simplifies to $64x^2 - 800x + 2275 = 0,$ which factors as $(8x - 65)(8x - 35) = 0.$  Since $x \le 5,$ $x = \frac{35}{8}.$  Then
\[\frac{(35/8)^2}{25} + \frac{y^2}{9} = 1.\]This leads to $y^2 = \frac{135}{64},$ so $y = \frac{\sqrt{135}}{8} = \pm \frac{3 \sqrt{15}}{8}.$  We can take $y = \frac{3 \sqrt{15}}{8}.$

Thus, the slope of line $AB$ is
\[\frac{\frac{3 \sqrt{15}}{8}}{\frac{35}{8} - 4} = \sqrt{15},\]so its equation is
\[y = \sqrt{15} (x - 4).\]To find $B,$ we substitute into the equation of the ellipse, to get
\[\frac{x^2}{25} + \frac{15 (x - 4)^2}{9} = 1.\]This simplifies to $128x^2 - 1000x + 1925 = 0.$  We could try factoring it, but we know that $x = \frac{35}{8}$ is a solution (because we are solving for the intersection of the line and the ellipse, and $A$ is an intersection point.)  Hence, by Vieta's formulas, the other solution is
\[x = \frac{1000}{128} - \frac{35}{8} = \frac{55}{16}.\]Then $y = \sqrt{15} (x - 4) = -\frac{9 \sqrt{15}}{16}.$  Hence,
\[BF = \sqrt{ \left( \frac{55}{16} - 4 \right)^2 + \left( -\frac{9 \sqrt{15}}{16} \right)^2} = \boxed{\frac{9}{4}}.\]